Refer back to the pizza-delivery Try It exercise. The mean delivery time is $36$minutes and the population standard deviation is six minutes. Assume the sample size is changed to $50$restaurants with the same sample mean.Find a $90$% confidence interval estimate for the population mean delivery time.

Given in the question that, the mean delivery time is $36$minutes and the population standard deviation is six minutes, the sample size is changed to $50$.

The sample mean is $36$minutes, that is $\stackrel{-}{x}$, =$36$

The random sample of $50$pizza delivery restaurants, that is, n=$50$

The standard deviation is $6$, that is $\sigma$, =$6$

Then calculate the confidence interval using Tl-$83$calculator using the steps given below,

First, press STAT and use the arrow (?) to TESTS.

Then, press the arrow localid="1650526391305" $(\downarrow )$to localid="1650526396394" $7$:localid="1650526399908" $Z$interval and press enter

Now, use the arrow localid="1650526404100" $(\to )$to STATS and press enter.

Then press the arrow (?) and enter the inputs as given below

Then press the arrow (?) to the calculator and press enter. The output is given below